【導(dǎo)讀】解:∵∠1+∠2=180°,∠1+∠EDF=180°,∴∠2=∠EDF,∴AB∥DF,∴∠3=∠AEF,又∵∠3=∠B,∴∠B=∠AEF,∴EF∥BC,∴∠ACB=∠AFE=50°∠ACB,交AB于E,∠DBF=∠F,問(wèn):CE∥DF嗎?∴2∠2=∠CAD+∠2,∴∠CAD=2∠2-∠2=∠2,即∠CAD=∠2,∴∠ACD=∠CAD,∵CA平分∠BCD,∴∠ACB=∠ACD,問(wèn)AD與BE平行嗎?=∠4,又∵∠1=∠2,∴∠2+∠CAE=∠4,即∠DAC=∠4,75°,EG平分∠AEC,試說(shuō)明AB∥EF∥CD.15°=75°,∵∠NCE=75°,∴∠CEF=∠NCE,∴EF∥CD,解:∠α+∠β=∠γ,理由:過(guò)點(diǎn)P作PQ∥l1,則∠CPQ=∠α,=∠γ,即∠α+∠β=∠γ若點(diǎn)P在線(xiàn)段AM上,則∠β-∠α=∠γ;